Optimal. Leaf size=104 \[ -\frac {a \left (2 a^2-3 b^2\right ) x}{2 b^4}+\frac {2 (a-b)^{3/2} (a+b)^{3/2} \text {ArcTan}\left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b^4}+\frac {\left (2 \left (a^2-b^2\right )-a b \cos (x)\right ) \sin (x)}{2 b^3}-\frac {\sin ^3(x)}{3 b} \]
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Rubi [A]
time = 0.18, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2774, 2944,
2814, 2738, 211} \begin {gather*} -\frac {a x \left (2 a^2-3 b^2\right )}{2 b^4}+\frac {\sin (x) \left (2 \left (a^2-b^2\right )-a b \cos (x)\right )}{2 b^3}+\frac {2 (a-b)^{3/2} (a+b)^{3/2} \text {ArcTan}\left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b^4}-\frac {\sin ^3(x)}{3 b} \end {gather*}
Antiderivative was successfully verified.
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Rule 211
Rule 2738
Rule 2774
Rule 2814
Rule 2944
Rubi steps
\begin {align*} \int \frac {\sin ^4(x)}{a+b \cos (x)} \, dx &=-\frac {\sin ^3(x)}{3 b}-\frac {\int \frac {(-b-a \cos (x)) \sin ^2(x)}{a+b \cos (x)} \, dx}{b}\\ &=\frac {\left (2 \left (a^2-b^2\right )-a b \cos (x)\right ) \sin (x)}{2 b^3}-\frac {\sin ^3(x)}{3 b}-\frac {\int \frac {b \left (a^2-2 b^2\right )+a \left (2 a^2-3 b^2\right ) \cos (x)}{a+b \cos (x)} \, dx}{2 b^3}\\ &=-\frac {a \left (2 a^2-3 b^2\right ) x}{2 b^4}+\frac {\left (2 \left (a^2-b^2\right )-a b \cos (x)\right ) \sin (x)}{2 b^3}-\frac {\sin ^3(x)}{3 b}+\frac {\left (a^2-b^2\right )^2 \int \frac {1}{a+b \cos (x)} \, dx}{b^4}\\ &=-\frac {a \left (2 a^2-3 b^2\right ) x}{2 b^4}+\frac {\left (2 \left (a^2-b^2\right )-a b \cos (x)\right ) \sin (x)}{2 b^3}-\frac {\sin ^3(x)}{3 b}+\frac {\left (2 \left (a^2-b^2\right )^2\right ) \text {Subst}\left (\int \frac {1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )}{b^4}\\ &=-\frac {a \left (2 a^2-3 b^2\right ) x}{2 b^4}+\frac {2 (a-b)^{3/2} (a+b)^{3/2} \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b^4}+\frac {\left (2 \left (a^2-b^2\right )-a b \cos (x)\right ) \sin (x)}{2 b^3}-\frac {\sin ^3(x)}{3 b}\\ \end {align*}
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Mathematica [A]
time = 0.22, size = 96, normalized size = 0.92 \begin {gather*} \frac {-12 a^3 x+18 a b^2 x-24 \left (-a^2+b^2\right )^{3/2} \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2}}\right )+3 b \left (4 a^2-5 b^2\right ) \sin (x)-3 a b^2 \sin (2 x)+b^3 \sin (3 x)}{12 b^4} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.17, size = 152, normalized size = 1.46
method | result | size |
default | \(-\frac {2 \left (\frac {\left (-a^{2} b -\frac {1}{2} b^{2} a +b^{3}\right ) \left (\tan ^{5}\left (\frac {x}{2}\right )\right )+\left (-2 a^{2} b +\frac {10}{3} b^{3}\right ) \left (\tan ^{3}\left (\frac {x}{2}\right )\right )+\left (-a^{2} b +b^{3}+\frac {1}{2} b^{2} a \right ) \tan \left (\frac {x}{2}\right )}{\left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}+\frac {a \left (2 a^{2}-3 b^{2}\right ) \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{2}\right )}{b^{4}}+\frac {2 \left (a -b \right )^{2} \left (a +b \right )^{2} \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {x}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b^{4} \sqrt {\left (a -b \right ) \left (a +b \right )}}\) | \(152\) |
risch | \(-\frac {a^{3} x}{b^{4}}+\frac {3 a x}{2 b^{2}}-\frac {i {\mathrm e}^{i x} a^{2}}{2 b^{3}}+\frac {5 i {\mathrm e}^{i x}}{8 b}+\frac {i {\mathrm e}^{-i x} a^{2}}{2 b^{3}}-\frac {5 i {\mathrm e}^{-i x}}{8 b}+\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i x}-\frac {i \sqrt {-a^{2}+b^{2}}-a}{b}\right ) a^{2}}{b^{4}}-\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i x}-\frac {i \sqrt {-a^{2}+b^{2}}-a}{b}\right )}{b^{2}}-\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i x}+\frac {i \sqrt {-a^{2}+b^{2}}+a}{b}\right ) a^{2}}{b^{4}}+\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i x}+\frac {i \sqrt {-a^{2}+b^{2}}+a}{b}\right )}{b^{2}}+\frac {\sin \left (3 x \right )}{12 b}-\frac {a \sin \left (2 x \right )}{4 b^{2}}\) | \(269\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.40, size = 243, normalized size = 2.34 \begin {gather*} \left [-\frac {3 \, {\left (a^{2} - b^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (x\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (x\right ) + b\right )} \sin \left (x\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (x\right )^{2} + 2 \, a b \cos \left (x\right ) + a^{2}}\right ) + 3 \, {\left (2 \, a^{3} - 3 \, a b^{2}\right )} x - {\left (2 \, b^{3} \cos \left (x\right )^{2} - 3 \, a b^{2} \cos \left (x\right ) + 6 \, a^{2} b - 8 \, b^{3}\right )} \sin \left (x\right )}{6 \, b^{4}}, \frac {6 \, {\left (a^{2} - b^{2}\right )}^{\frac {3}{2}} \arctan \left (-\frac {a \cos \left (x\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (x\right )}\right ) - 3 \, {\left (2 \, a^{3} - 3 \, a b^{2}\right )} x + {\left (2 \, b^{3} \cos \left (x\right )^{2} - 3 \, a b^{2} \cos \left (x\right ) + 6 \, a^{2} b - 8 \, b^{3}\right )} \sin \left (x\right )}{6 \, b^{4}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 194 vs.
\(2 (86) = 172\).
time = 0.51, size = 194, normalized size = 1.87 \begin {gather*} -\frac {{\left (2 \, a^{3} - 3 \, a b^{2}\right )} x}{2 \, b^{4}} - \frac {2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, x\right ) - b \tan \left (\frac {1}{2} \, x\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{4}} + \frac {6 \, a^{2} \tan \left (\frac {1}{2} \, x\right )^{5} + 3 \, a b \tan \left (\frac {1}{2} \, x\right )^{5} - 6 \, b^{2} \tan \left (\frac {1}{2} \, x\right )^{5} + 12 \, a^{2} \tan \left (\frac {1}{2} \, x\right )^{3} - 20 \, b^{2} \tan \left (\frac {1}{2} \, x\right )^{3} + 6 \, a^{2} \tan \left (\frac {1}{2} \, x\right ) - 3 \, a b \tan \left (\frac {1}{2} \, x\right ) - 6 \, b^{2} \tan \left (\frac {1}{2} \, x\right )}{3 \, {\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}^{3} b^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.11, size = 1677, normalized size = 16.12 \begin {gather*} \frac {\frac {4\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (3\,a^2-5\,b^2\right )}{3\,b^3}-\frac {\mathrm {tan}\left (\frac {x}{2}\right )\,\left (-2\,a^2+a\,b+2\,b^2\right )}{b^3}+\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^5\,\left (2\,a^2+a\,b-2\,b^2\right )}{b^3}}{{\mathrm {tan}\left (\frac {x}{2}\right )}^6+3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+1}-\frac {2\,\mathrm {atanh}\left (\frac {64\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}}{128\,a\,b^2+112\,a^2\,b-352\,a^3-64\,b^3+\frac {16\,a^4}{b}+\frac {320\,a^5}{b^2}-\frac {112\,a^6}{b^3}-\frac {96\,a^7}{b^4}+\frac {48\,a^8}{b^5}}+\frac {144\,a^2\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}}{128\,a\,b^4+16\,a^4\,b+320\,a^5-64\,b^5+112\,a^2\,b^3-352\,a^3\,b^2-\frac {112\,a^6}{b}-\frac {96\,a^7}{b^2}+\frac {48\,a^8}{b^3}}+\frac {80\,a^3\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}}{128\,a\,b^5+320\,a^5\,b-112\,a^6-64\,b^6+112\,a^2\,b^4-352\,a^3\,b^3+16\,a^4\,b^2-\frac {96\,a^7}{b}+\frac {48\,a^8}{b^2}}-\frac {144\,a^4\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}}{128\,a\,b^6-112\,a^6\,b-96\,a^7-64\,b^7+112\,a^2\,b^5-352\,a^3\,b^4+16\,a^4\,b^3+320\,a^5\,b^2+\frac {48\,a^8}{b}}+\frac {48\,a^5\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}}{48\,a^8-96\,a^7\,b-112\,a^6\,b^2+320\,a^5\,b^3+16\,a^4\,b^4-352\,a^3\,b^5+112\,a^2\,b^6+128\,a\,b^7-64\,b^8}-\frac {192\,a\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}}{128\,a\,b^3-352\,a^3\,b+16\,a^4-64\,b^4+112\,a^2\,b^2+\frac {320\,a^5}{b}-\frac {112\,a^6}{b^2}-\frac {96\,a^7}{b^3}+\frac {48\,a^8}{b^4}}\right )\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}}{b^4}+\frac {a\,\mathrm {atan}\left (\frac {\frac {a\,\left (2\,a^2-3\,b^2\right )\,\left (\frac {8\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (8\,a^9-16\,a^8\,b-16\,a^7\,b^2+48\,a^6\,b^3-3\,a^5\,b^4-39\,a^4\,b^5+11\,a^3\,b^6+7\,a^2\,b^7+4\,a\,b^8-4\,b^9\right )}{b^6}-\frac {a\,\left (\frac {8\,\left (4\,a^5\,b^8-6\,a^4\,b^9-6\,a^3\,b^{10}+10\,a^2\,b^{11}+2\,a\,b^{12}-4\,b^{13}\right )}{b^9}-\frac {a\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (2\,a^2-3\,b^2\right )\,\left (8\,a^3\,b^8-16\,a^2\,b^9+8\,a\,b^{10}\right )\,4{}\mathrm {i}}{b^{10}}\right )\,\left (2\,a^2-3\,b^2\right )\,1{}\mathrm {i}}{2\,b^4}\right )}{2\,b^4}+\frac {a\,\left (2\,a^2-3\,b^2\right )\,\left (\frac {8\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (8\,a^9-16\,a^8\,b-16\,a^7\,b^2+48\,a^6\,b^3-3\,a^5\,b^4-39\,a^4\,b^5+11\,a^3\,b^6+7\,a^2\,b^7+4\,a\,b^8-4\,b^9\right )}{b^6}+\frac {a\,\left (\frac {8\,\left (4\,a^5\,b^8-6\,a^4\,b^9-6\,a^3\,b^{10}+10\,a^2\,b^{11}+2\,a\,b^{12}-4\,b^{13}\right )}{b^9}+\frac {a\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (2\,a^2-3\,b^2\right )\,\left (8\,a^3\,b^8-16\,a^2\,b^9+8\,a\,b^{10}\right )\,4{}\mathrm {i}}{b^{10}}\right )\,\left (2\,a^2-3\,b^2\right )\,1{}\mathrm {i}}{2\,b^4}\right )}{2\,b^4}}{\frac {16\,\left (-4\,a^{11}+6\,a^{10}\,b+18\,a^9\,b^2-31\,a^8\,b^3-26\,a^7\,b^4+59\,a^6\,b^5+8\,a^5\,b^6-49\,a^4\,b^7+10\,a^3\,b^8+15\,a^2\,b^9-6\,a\,b^{10}\right )}{b^9}+\frac {a\,\left (2\,a^2-3\,b^2\right )\,\left (\frac {8\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (8\,a^9-16\,a^8\,b-16\,a^7\,b^2+48\,a^6\,b^3-3\,a^5\,b^4-39\,a^4\,b^5+11\,a^3\,b^6+7\,a^2\,b^7+4\,a\,b^8-4\,b^9\right )}{b^6}-\frac {a\,\left (\frac {8\,\left (4\,a^5\,b^8-6\,a^4\,b^9-6\,a^3\,b^{10}+10\,a^2\,b^{11}+2\,a\,b^{12}-4\,b^{13}\right )}{b^9}-\frac {a\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (2\,a^2-3\,b^2\right )\,\left (8\,a^3\,b^8-16\,a^2\,b^9+8\,a\,b^{10}\right )\,4{}\mathrm {i}}{b^{10}}\right )\,\left (2\,a^2-3\,b^2\right )\,1{}\mathrm {i}}{2\,b^4}\right )\,1{}\mathrm {i}}{2\,b^4}-\frac {a\,\left (2\,a^2-3\,b^2\right )\,\left (\frac {8\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (8\,a^9-16\,a^8\,b-16\,a^7\,b^2+48\,a^6\,b^3-3\,a^5\,b^4-39\,a^4\,b^5+11\,a^3\,b^6+7\,a^2\,b^7+4\,a\,b^8-4\,b^9\right )}{b^6}+\frac {a\,\left (\frac {8\,\left (4\,a^5\,b^8-6\,a^4\,b^9-6\,a^3\,b^{10}+10\,a^2\,b^{11}+2\,a\,b^{12}-4\,b^{13}\right )}{b^9}+\frac {a\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (2\,a^2-3\,b^2\right )\,\left (8\,a^3\,b^8-16\,a^2\,b^9+8\,a\,b^{10}\right )\,4{}\mathrm {i}}{b^{10}}\right )\,\left (2\,a^2-3\,b^2\right )\,1{}\mathrm {i}}{2\,b^4}\right )\,1{}\mathrm {i}}{2\,b^4}}\right )\,\left (2\,a^2-3\,b^2\right )}{b^4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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